Problem - 3249
Compute $$\sqrt{\frac{2}{2^2}+\sqrt{\frac{2}{2^4}+\sqrt{\frac{2}{2^8}+\cdots}}}$$
First, it can be show that (using the same technique as that in # 3882):
$$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=2$$
Then the given expression equals
$$\sqrt{\frac{1}{2^2}}\times\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=\boxed{1}$$